Suppose that we have two smooth functions from some real manifold $X$ into the complex plane $\mathbb C$:

$$f,g: X \to \mathbb C.$$

I'm vaguely wondering, if we treat $\mathbb C$ as a plane equipped with a flat metric, if there's a relationship between (complex) multiplication and the Frechet derivative map, i.e. a kind of "product rule" which relates $h(x):= f(x)g(x)$ to $f$ and $g$.

1 Answer
1

Since $\mathbb{C}$ is a vector space, its tangent spaces can be canonically identified with itself. Hence, for each $x\in X$, we may view $D_xf$ as a map
$$D_xf:T_xX\to\mathbb{C}.$$
Then,
$$D_x(fg)=f(x)D_xg+g(x)D_xf.$$
To prove this formula, note that $fg=m\circ(f\times g)$ where $m:\mathbb{C}\times\mathbb{C}\to\mathbb{C}$ is multiplication and $f\times g:X\to\mathbb{C}\times\mathbb{C}$ is the map $(f\times g)(x)=(f(x),g(x))$. Then, $D_x(f\times g)=D_xf\times D_xg:T_xX\to\mathbb{C}\times\mathbb{C}$. Moreover, after identifying $T_z\mathbb{C}$ with $\mathbb{C}$ for $z\in\mathbb{C}$ we have that for all $(z_1,z_2)\in\mathbb{C}\times\mathbb{C}$,
$$D_{(z_1,z_2)}m:\mathbb{C}\times\mathbb{C}\to\mathbb{C},\quad (a,b)\mapsto z_1b+z_2a$$
since
$$(D_{(z_1,z_2)}m)(a,b)=\left.\frac{d}{dt}\right|_{t=0}(z_1+at)(z_2+bt)=z_1b+z_2a.$$
Combining this with the chain rule, we get
$$D_x(fg)=D_x(m\circ(f\times g))=D_{(f(x),g(x))}m\circ(D_xf\times D_xg)=f(x)D_xg+g(x)D_xf.$$

$\begingroup$Marvellous! I believe the canonical identification (plus inexperience) was what was tripping me up there. That does seem jolly reasonable.$\endgroup$
– BenAug 7 at 15:45

$\begingroup$@Ben My pleasure. By the way, if $V$ is a vector space and $v\in V$, the canonical identification between $V$ and $T_vV$ is the linear isomorphism $V\to T_vV$ which sends a vector $w$ to the derivation $C^\infty(V)\to\mathbb{R}:f\mapsto\left.\frac{d}{dt}\right|_{t=0}f(v+tw)$.$\endgroup$
– SpenserAug 7 at 15:47

$\begingroup$Of course! That makes a lot of sense$\endgroup$
– BenAug 7 at 16:33